Lesson 9 • X-rays

– Roentgen’s Early Experiment – How are X-rays Produced – Hard and Soft X-RAYS – X-ray Spectra – X-ray spectra according to Bohr’s Theory – Properties of X Rays – Uses of X Rays – Dangers of X Rays – Tutorials

• Photoelectron Effect • Energy Levels

Roentgen’s Early Experiment • Wilhelm K. Roentgen, who in 1901 was the first man to receive the

Nobel Prize in physics, first observed x rays in 1895. He was studying the light produced when electricity was passed through a gas in a tube at low pressure.

• He noted that a paper screen coated with a fluorescent material glowed when it was in the vicinity of the tube under operation. We now know that x rays are produced if electrons are accelerated through a potential of the order of 104 or more volts and then allowed to strike a metal target.

• Classical electromagnetic theory indicates that the deceleration of an electric charge causes it to radiate energy. In this case the form of electromagnetic radiation is called x rays.

• Early observations showed that this newly detected radiation had greater penetration power than any other electromagnetic radiation known at that time. It was also observed that these x rays affected a photographic film and would ionize atoms.

• These effects are utilized in detecting devices for x rays. Since x rays are electromagnetic waves, they are not deflected by either an electric or a magnetic field.

Roentgen’s Early Experiment

How are X-rays Produced? Filament X-Ray Tube Invented by Coolidge in 1913. The most widely-used laboratory X-Ray source. Major components are a water-cooled target (anode) and a tungsten filament (cathode) that emits electrons. A high potential (up to 60 kV) is maintained between the filament and the anode, accelerating the electrons into the anode and generating X-Rays. Cooling water is circulated through the anode to keep it from melting (>99% of input power generates heat). Interior of the tube is evacuated for the electron beam; thin beryllium windows transmit the X-Rays

Ceramic Diffraction X-Ray Tube Schematic View

How are X-rays Produced?

Schematic diagram of x-ray tube and circuit

• Filament is heated, releasing electrons via thermionic emission (Vf ~ 10V, If ~ 4A, resulting in T>2000oC) • X rays are produced by high-speed electrons bombarding the target • Typically < 1% of energy is converted to x rays; the rest is heat

How are X-rays Produced?

• The electron beam produced and controlled by the current that is passed through the filament.

• Stable high voltage and filament current power supplies are needed (old-style transformers →high frequency supplies).

• Power rating: applied potential ×electron beam current (example: 50 kV and 40 mA →2 kW).

• Maximum power determined by the rate of heat removal (without water, a tube can be destroyed in seconds → flow interlocks).

• The anode is electrically grounded, while the filament is kept at negative kV’s (the water-cooled anode won’t short out, and the filament is protected by glass insulation).

How are X-rays Produced?

Hard and Soft X-RAYS

• The wavelength of the x rays is controlled by the applied voltage between the cathode and anode. For the higher potential differences (short wavelengths) the term hard x rays is used and for the lower potential differences (long wavelengths) the term soft x rays is used to describe the quality of the radiation.

• Since the tube is highly evacuated, the electron beam current (usually in the 10 milliampere range) is determined by the filament current. When these electrons strike the metal anode target some of them generate x rays as they are abruptly brought to rest.

• This deceleration radiation is called Bremsstrahlung (braking radiation) which appears as the continuous x-ray background shown in Figure

X-ray Spectra

The Bremsstrahlung Spectrum

X-ray Spectra

• Much of the electron energy goes into heating up the anode. Some of the electrons in the beam interact with the innermost, most tightly, bound electrons in the target and "knock" them into excited states. When these excited atoms return to their ground state, photons are emitted. Since the target material is a metal with many electrons, the innermost electron energy levels are of the order of thousands of electron volts (keV).

• The photons emitted due to this excitation are characteristic of the target material and are called the characteristic spectra .

• The maximum energy x rays correspond to the conversion of the maximum electron beam energy into a photon, electron beam energy = photon energy maximum

• eV = hfmax = hc/ λmin (30.1) • where V is the accelerating voltage for the tube, fmax is

maximum x-ray frequency, h is Planck's constant, and e is the change of the electron.

X-ray Spectra • The x-ray photons interact with matter through the

photoelectric effect and Compton scattering. The photoelectric effect increases rapidly with Z, the atomic number of the target material. The photoelectric interaction probability is proportional to Z and is the greatest for low-energy photons.

• The Compton effect is relatively independent of energy and the atomic number of the absorber. These two processes determine the absorption coefficient of the material for x rays.

• The intensity x rays passing through a material of thickness x can be expressed as an exponential function,

I is the x-ray intensity at a distance x in the material, µ is absorption coefficient of the material, Δx is the thickness of the material, and IO is the incident intensity of the x rays in joules per unit area per second.

Properties of X Rays i. Like radio waves and other electromagnetic radiation, X rays have a wide range of

wavelengths (Obey the relationship . Strong, deeply penetrating, and highly destructive rays with short wavelengths are called hard X rays. Those with longer wavelengths and less penetrating power—the type used in medical and dental diagnosis—are known as soft X rays.

ii. X rays can penetrate some substances more easily than others. For example, they penetrate flesh more easily than bone, and bone more easily than lead. Thus they make it possible to see bones within flesh and a bullet embedded in bone. The ability of X rays to penetrate depends not only on their wavelength, but also on the density and thickness of the substance.

iii. X rays affect photographic film in the same way as light rays do. An X-ray photograph is made by passing a beam of X rays through the subject onto photographic film. In a more recent technique, called xeroradiography, an electrostatic ally charged metal plate is substituted for photographic film. When X rays pass through an object and strike the plate, they discharge it in proportion to the density of the object. To bring out the electrostatic image thus formed, the plate is sprayed with a powder that adheres to the charged area. The powder gathers more thickly in heavily charged areas than in lightly charged spots, producing a detailed picture.

iv. X rays cause certain substances to glow, or fluoresce.

Properties of X Rays

• Bragg's Law • According to W. L. Bragg, X-ray diffraction can be viewed as a process that is

similar to reflection from planes of atoms in the crystal. In Bragg's construct, the planes in the crystal are exposed to a radiation source at a glancing angle θ and X rays are scattered with an angle of reflection also equal to θ. The incident and diffracted rays are in the same plane as the normal to the crystal planes.

• Constructive interference occurs only when the path length difference between rays scattered from parallel crystal planes would be an integer number of wavelengths of the radiation. When the crystal planes are separated by a distance d, the path length difference would be 2dsin θ.

Properties of X Rays • Thus, for constructive interference to occur the following relation must

hold true. • n λ = 2 d sin θ

• This relation is known as Bragg's Law. Thus for a given d spacing and wavelength, the first order peak (n = 1) will occur at a particular θ value. Similarly, the θ values for the second (n = 2) and higher order (n > 2) peaks can be predicted.

• The above derivation assumes that phase differences between wavelets scattered at different points depend only on path length differences.

• It is assumed that there is no intrinsic phase change between the incident and scattered beams or that this phase change is constant for all scattering events. However this is not always the case (see anomalous scattering elsewhere).

wavelengthBrogliedetheiseVm

hwhere

eVm

nh

m

eVm

hndthen

mv

hso

hmvpand

m

eVv

hpwheremvPAndeVmvFrom

222sin2

2

,2

1 2

Properties of X Rays

Uses of X Rays

• In Medicine • X rays help dentists detect diseases of the teeth. Doctors use X rays to

locate bullets and other foreign objects within the body; to guide them in setting broken bones; and to detect cancer, ulcers, kidney stones, and other abnormalities.

• Various types of X-ray scanners have been developed that allow highly detailed views of a particular section of the body. One type, known as a CT (computerized tomography) scanner, sends narrow beams of X rays at various angles through a patient's body. The information obtained from the X rays is processed by a computer to produce an image of a cross-section of the body. The image shows much more detail than an ordinary X-ray picture. A section of the body can be studied in three dimensions by producing a series of adjacent cross-sectional images.

• X rays can halt the growth of cells and even destroy them altogether. They are therefore used to destroy benign and malignant tumors. X rays have also been used in the treatment of leukemia and bursitis.

Uses of X Rays

• In Industry • X rays are used to inspect canned goods and other

packaged products. A conveyor carries the goods past a beam of X rays. If a container is improperly filled, or if it contains a foreign substance, the X rays set off an alarm or set into action a device that removes the container from the conveyor. X rays are similarly used to separate beryl from granite and to inspect airplane and automobile parts, rubber goods, plastics, metal castings, and a variety of other products.

• When used as a target in an X-ray tube, every element gives off X rays of specific wavelengths. These characteristic rays are used to analyze metal alloys, paint pigments, and other substances.

Uses of X Rays

• In Science (crystallography) • Scientists have learned much about the structure of matter

by means of X rays. Among other things, they have learned how atoms are arranged in crystals. The average wavelength of X- rays is about equal to the distance between the atoms in crystals. Crystals therefore act as diffraction gratings for X rays. That is, they scatter X- rays in a pattern that shows the positions of their atoms.

• When the patterns of specific crystalline substances are known, technicians can use X- rays to analyze substances of which they are a part. Petroleum products, metal alloys, and other substances are thus analyzed.

Uses of X Rays

• Other Uses • Airport security and other personnel use X- rays in

examining luggage and packages to check for weapons or smuggled articles.

• X-rays show whether pearls are natural or cultured, and whether gemstones are natural or synthetic.

• X-rays have also been used to learn whether paintings attributed to noted painters are authentic. Sometimes they have revealed changes made in the original work, or an earlier painting under the one that appears on the surface.

• In geology labs, the most common targets used are copper and cobalt.

Dangers of X- Rays

• Because X rays can kill living cells, they must be used with extreme care. When improperly used they can cause severe burns, cancer, leukemia, and cataracts. They can speed aging, reduce immunity to disease, and bring about disastrous changes in the reproductive cells.

• Lead screens, sheets of lead-impregnated rubber, and leaded glass are used to shield patients and technicians from undesired radiation.

• The effect of X radiation is cumulative. That is, a number of minor doses over a number of years is equivalent to a large dose at one time.

Tutorials

1. Electrons are accelerated from rest through a p.d of 10Kv in an x ray tube. Calculate:

(i) The resultant energy of the electrons in eV. (104ev)

(ii) The wavelength of the associated electron waves. (1.23x10-11

m)

(iii) The maximum energy and the minimum wavelength of the x ray radiation

generated (assume, me,e, h=6.62x10-34

Js, c=3x108m/s): (1.6x10

-15J, 1.24x10

-10m)

Hint:use the de Broglie equation

mxeV

hcand

eVm

h

eVmpwhereP

he

10

min 1024.1

2

2

2. The energy of an x ray photon is hf. X rays are emitted from a target bombarded by

electrons which have been accelerated from rest through 105V. Calculate the minimum

possible wavelength of the X rays assuming that the corresponding energy is equal to the

whole of the kinetic energy of one electron. ( assume h, e, c ) (use: hf = eV),

ans, 1.243x10-11

m

Tutorials

3. a) State the de Broglie relationship in words.

b) Show that the speed of electrons which have been accelerated from rest through a p.d

of 2Kv is 2.6x107m/s and calculate the wavelength with a beam of these electrons.

(2.7x10-11

m) .

c) Suggest why electron diffraction is a useful tool for studying the arrangement of atoms

in a crystals where the separation of atoms is typically 0.3nm (λ is less than 0.03nm so

diffraction is observed)

4. Electrons are accelerated through a p.d of 50V. Calculate,

(a) The speed acquired.(4.123x106m/s)

(b) Their momentum (P = mv)(37.52x10-25

kgm/s)

(c) The de Broglie wavelength associated with them. (h/p = 0.17nm)

Tutorials

5. What is the k.e of an electron with a de Broglie wavelength of 0.1nm.through what p.d

should it be accelerated to achieve this value? Assume e, me, h.

(use

KvVem

hVsoekeValso

m

hvce

m

hmmvekbutmvp

m

hv

vm

h

p

h

e

e

ee

e

151.0,2

,.

sin

2

1

2

1.,

2

2

22

22

22

22

2

)

6. An x-ray operates at 30Kv and the current through it is 2.0mA.Calulate:

(i) The electrical power output

(ii) The number of electrons striking the target per second.

(iii)The speed of the electrons when they hit the target

(iv) The lower wavelength limit of the x-rays emitted.

Tutorials Solution.

(i)

mxeV

hc

s

mx

x

xx

m

eVv

eVmviii

xx

x

e

In

ondperettthestrickingelectronsof

numbertheisnwhereneIii

w

xxxVIpi

10

min

8

31

16

2

16

19

3

33

1041.0

101109

106.122

2

1)(

103.1106.1

102

secarg

)(

60

100.21030)(

7. Monochromatic X-rays of wavelength 1.2x10-10

m are incident on a crystal. The ist order

diffraction maximum is observed at when the angle between the incident beam and the

atomic plane is 120

(i) What is the separation of the atomic planes responsible for the diffraction?

(d = 2.89x10-10

m)

(ii) What is the highest order Bragg diffraction observable? ( max n = 4)

Tutorials

8. An x-ray machine can accelerate electrons of energies 4.8x10-15J. The shortest

wavelength of the x- rays produced by the machine is found to be 4.1x10-11m.

Use this information to estimate the value of the plank constant. ( use,

Jsxhhc

hfE 34

min

1056.6,

9. Molybdemumka x-rays have wavelength 7x10-11m. Find:

(i) The minimum x ray potential difference that can produce these x rays. (17.7Kv)

(ii) Their photon energy in e V (17.7 Kv x e = 17.7KeV)

10. The spacing between Principal planes of Nacl crystal is 2.82Å . It is found that the first

order Bragg diffraction occurs at an angle of 100. What is the wavelength of the x rays?

mxnd

nd

11108.9/sin2

sin2

Tutorials

11. Bragg’s spectrometer is set for the first order reflection to be received by the dectector at

a glancing angle of 9.80. calculate the angle through which the detector is rotated to

receive 2nd

order reflection from the same face of the crystal.

2sin23.9sin2

sec

sin2sin2

2

2121

dandd

orderondandfirsttheforanglesglamcing

areandwheredandd

This gives θ2= 18.80 and therefore

0

12 5.93.98.18

Photoelectron Effect

The photoelectric effect relates to the following phenomena: if a metal surface is illuminated by visible or ultraviolet light radiation, electrons are released provided that the frequency of the radiation exceeds a critical threshold. In 1902 Lernard found that the velocity or kinetic energy of the electron emitted from an illuminated metal was independent of the intensity of the particular incident monochromatic light. It appeared to vary only with the wavelength or the frequency of the incident light. Quantum Theory of Radiation, Planck’s Constant In 1902 Planck had shown that the experimental observations in black body heat radiation could be explained on the basis that the energy from the body was emitted in separate packets of energy. Each packet was called quantum of energy and the amount

of energy E carried is given by Where h = 6.63 × 10-34 Js called the Planck’s constant and f is the frequency of

radiation

E =hf

Photoelectron Effect

Work Function The least or minimum amount of work or energy necessary to take free electron out of a metal

against the attractive forces of surrounding positive ions is called the work function of the metal,

symbol w . The work functions of cesium, sodium and beryllium are respectively about 1.9 eV,

2.0 eV and 3.9 eV . Electron volt (eV) is a unit of energy equal to 1 electron charge e × 1 volt,

which is

Hence for example the work of sodium, w =2eV = 2×1.6×10-19

J = 3.2×10-19

J.

Einstein’s Particle (Photon) Theory

In 1905 Einstein suggested that the experimental results in photoelectricity could be explained by

applying a quantum theory of light. He assumed that light of frequency f contains packets or

quanta of energy hf. On this basis, light consists of particles called photons. The number of

photons per unit area of cross section of the beam of light per second is proportional to its

intensity. But the energy of a photon is proportional to its frequency and is independent of the

light intensity.

1.6 × 10-19

C × 1V = 1.6 × 10-19

J

Photoelectron Effect

Considering the case of sodium whose work function is 2eV or 3.2×10-19

J. According to

the particle theory, if the quantum energy in the incident light is 3.2×10-19

J, then the

electrons are just liberated from the metal. The particular frequency f is called the

threshold frequency of the metal. The threshold wavelength /c f . Hence the

threshold frequency f and threshold for sodium is E=hf w ,

19

14

34

3.2 104.8 10 Hz

6.6 10

wf

h

87

14

3 106.2 10 m

4.8 10

c

f

So electrons are not liberated from sodium if the incident light has a frequency less than 4.8 ×

1014

Hz or a wavelength longer than 6.2 × 10-7

m.

Photoelectron Effect

Einstein’s Photoelectric Equation

If the quantum of energy hf in the light incident on a metal is say 4.2 × 10-19

J, and the work

function wo of the metal is 3.2 × 10-19

J, the maximum energy of the liberated electrons is (4.2 ×

10-19

J - 3.2 × 10-19

J) = 1.0 × 10-19

J, because wo is the least energy to liberate electrons from the

metal. The maximum kinetic energy of the liberated electrons is given by2

max m

1E or v

2em .

The above equations are called the Einstein’s photoelectric equations.

max

max

E

E

hf w

hf w

Photoelectron Effect

Example

Calculate the maximum energy and kinetic energy of emitted electron when sodium is

illuminated by radiation of wavelength 150 nm. (wo for sodium is 2.0eV)

Photon energy is given by E =

34 819

9

6.63 10 3 1013.2 10 J

150 10

chf h

So maximum kinetic energy =19 19 -1813.2 10 2 1.6 10 J=10 Jhf w

The threshold frequency is given by

1914

-34

2 1.6 10 J4.8 10 Hz

6.63 10 Js

wf

h

Photoelectron Effect

Stopping Potential

Stopping potential is defined as the negative potential difference (p.d.) which can stop liberated

electrons with maximum energy. Stopping potential is denoted by Vs, hence

max sE =eV

hchf w w

Example

Caesium has a work function of 1.9eV. Find i) its threshold wavelength; ii) maximum energy of

liberated electrons when th metal is illuminated by light of wavelength 4.5 × 10-7

m; iii) stopping

potential. (take 1eV = 1.6×10-19

J, c = 3×108m/s, h = 6.63×10

-34Js).

i) Threshold wavelength /

c c ch

f w h w recall that E=hf w

8 347

19

3 10 6.63 106.5 10 m

1.9 1.6 10

Photoelectron Effect

ii) Maximum energy of liberated electrons =hf-wo where /f c hence

Max. Energy

34 819 19

7

6.63 10 3 101.9 10 1.4 10 J

4.5 10

hcw

iii) the stopping potential Vs is given by eVs = 1.4×10-19

J where e = 1.6×10-19

C recall Emax=eVs

therefore

-19

-19

1.4 10V= 0.9V (approx)

1.6 10

Homework

1 Light of frequency 5.0×1014

Hz liberates electrons with energy 2.31×10-19

J from a certain

metallic surface. What is the wavelength of ultraviolet light which liberates electrons of

energy 8.93×10-19

J from the same surface? Take c =3.0×108m/s, h=6.63×10

-34Js.

2 When light of frequency 5.4×1014

Hz is shone on to a metal surface the maximum energy

of the electrons emitted is 1.2×10-19

J. If the same surface is illuminated with light of

frequency 6.6×1014

Hz the maximum energy of the electrons emitted is 2.0×10-19

J. Use

this data to calculate a value for the Planck’s constant.